__Frequently Asked Questions on Chapter 4 – Quadratic Equations__

__Frequently Asked Questions on Chapter 4 – Quadratic Equations__

**Check whether the following are quadratic equations (x + 1) ^{2} = 2(x – 3)?**

(x + 1)^{2} = 2(x – 3) By using the formula for (a+b)^{2}=a^{2}+2ab+b^{2}*x*^{2} + 2*x* + 1 = 2*x* – 6 *x*^{2} + 7 Since the above equation is in the form of *ax*^{2} + *bx* + *c* = 0. Therefore, the given equation is a quadratic equation.

Quadratic Equations

**Find the roots of the following quadratic equations by factorisation x^{2} – 3x – 10 = 0.**

Given, *x ^{2}* – 3

*x*– 10 =0 Taking LHS,

*x*

^{2}– 5

*x*+ 2

*x*– 10

*x*(

*x*– 5) + 2(

*x*– 5) (

*x*– 5)(

*x*+ 2) The roots of this equation,

*x*

^{2}– 3

*x*– 10 =0 are the values of x for which (

*x*– 5)(

*x*+ 2) = 0 Therefore,

*x*– 5 = 0 or

*x*+ 2 = 0

*x*= 5 or

*x*= -2 No, not every chemical reaction is a redox reaction. Reactions like double decompositions, acid-base neutralisation reactions, precipitation reactions are non-redox reactions.

Quadratic Equations

**Find the roots of the following quadratic equations, if they exist, by the method of completing the square 4 x^{2} +** 4√3

*x*+ 3 = 0.

4*x*^{2} + 4√3*x* + 3 = 0 Converting the equation into a^{2}+2ab+b^{2 }form, we get, (2*x*)^{2} + 2 × 2*x* × √3 + (√3)^{2} = 0 (2*x* + √3)^{2} = 0 (2*x* + √3) = 0 and (2*x* + √3) = 0 Therefore, either *x* = -√3/2 or *x* = -√3/2.

Quadratic Equations

**Find the nature of the roots of the following quadratic equations. If the real roots exist, find them 2 x^{2} – 3x + 5 = 0.**

Given,*2x*^{2} – 3*x* + 5 = 0 Comparing the equation with *ax*^{2} + *bx *+ *c* = 0, we get*a* = 2, *b* = -3 and *c* = 5 We know, Discriminant = *b*^{2} – 4*ac**= *( – 3)^{2} – 4 (2) (5) = 9 – 40 = – 31 As you can see, b^{2} – 4ac < 0 Therefore, no real root is possible for the given equation, *2x*^{2} – 3*x* + 5 = 0.

Quadratic Equations

**Find the values of k for each of the following quadratic equations, so that they have two equal roots 2x^{2} + kx + 3 =** 0.

2*x*^{2} + *kx* + 3 = 0 Comparing the given equation with *ax*^{2} + *bx *+ *c* = 0, we get,*a* = 2, *b* = k and *c* = 3 As we know, Discriminant = *b*^{2} – 4*ac*

= (*k*)^{2} – 4(2) (3) = *k*^{2} – 24 For equal roots, we know, Discriminant = 0*k*^{2} – 24 = 0*k*^{2} = 24 k = ±√24 = ±2√6

Quadratic Equations

**Check whether the equation (x + 1)^2 = 2(x – 3) is a quadratic equation**

(x + 1)^2 = 2(x – 3)

By using the formula (a+b)^2 = a^2 + 2ab + b^2

x^2 + 2x + 1 = 2x – 6

x^2 + 7 = 0

Since the above equation is in the form of ax^2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.

By factorisation method find the roots for the x^2 – 3x – 10 = 0 quadratic equation

x^2 – 3x – 10 = 0

Let us consider LHS,

x^2 – 5x + 2x – 10

x(x – 5) + 2(x – 5)

(x – 5) (x + 2)

x – 5 = 0 or x + 2 = 0

x = 5 or x = -2

The roots of the equation, x^2 – 3x – 10 = 0 are the values of x for which (x – 5) (x + 2) = Therefore, x = 5 or x = -2

**Find the two sides if the altitude of a right triangle is 7 cm less than its base. And if the hypotenuse is 13 cm.**

Given, the altitude of right triangle = (x – 7) cm

Let us consider, the base of the right triangle be ‘x’ cm.

From Pythagoras theorem, we know,

Base^2 + Altitude^2 = Hypotenuse^2

x^2 + (x – 7)^2 = 13^2

x^2 + x^2 + 49 – 14x = 169

2x^2 – 14x – 120 = 0

x^2 – 7x – 60 = 0

x^2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12) (x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

x = 12 or x = – 5

Since sides cannot be negative, x = 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.